# The image of a map of separable algebras

I'm reading Lenstra's notes on the étale fundamental group, and I've got stuck on his exercise 3.9(c). He says that if $A$ and $B$ are separable algebras over a field $K$, and $f:A \to B$ a $K$-algebra homomorphism, then the image of $f$ is precisely $\{b \in B: b \otimes 1 = 1 \otimes b\text{ in }B \otimes_A B\}$.

By his theorem 2.7, every separable $K$-algebra is a finite product of separable extension fields $K_i$ of $K$. So the morphisms of two such algebras are fairly restricted: if I'm not mistaken, if $A = \prod K_i$ and $B = \prod L_j$, and $v_i$ and $w_j$ are the respective idempotents defining these splittings, then each $f(v_i)$ is a sum of distinct $w_j$, each generating an extension field of $K_i$.

I don't know how to get from this to the claim. More generally, for any map of commutative rings $A \to B$, we have this subring $\{b\in B:b \otimes 1 = 1 \otimes b \text{ in }B \otimes_A B\}$, which I feel like I've seen in other contexts as well, though I can't remember where. Is there a high-level way of thinking about this set, or more general conditions that tell you it's equal to $f(A)$? Thank you in advance.

• One approach: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally, which reduces you to the case that $A$ is simply a field, which makes the tensor product $B \otimes_A B$ less mysterious. Dec 26 '12 at 4:41

Let $A$ be a field, $f: A \rightarrow B$ a map of rings. I claim first that in this baby case $f(A) = \{b: b \otimes 1 = 1 \otimes b \in B \otimes_A B \}$. Indeed, we have the inclusion $\subseteq$, and in the other direction, if $b \notin f(A)$ we may extend $1,b$ to basis $b^i$ for $B$ over $A$, as $b^i \otimes b^j$ form a basis for $B \otimes_A B$, in particular we have $1 \otimes b \neq b \otimes 1$.
Let's reduce to this case for the question at hand: you have the natural inclusion of $A$ modules $f(A) \rightarrow \{b \in B: b \otimes 1 = 1 \otimes b\}$, to prove this is an isomorphism it suffices to do it locally (the latter is easily seen to be an $A$ module). Indeed, suppose $A = K_1 \oplus \ldots \oplus K_n$, the spectrum of $A$ is just $\mathfrak{p}_1 \ldots \mathfrak{p}_n$, where $\mathfrak{p}_i = \oplus_{j \neq i} K_j \subseteq A$.
Fix some $\mathfrak{p}_i =: p$. Now, $A_{p}$ is just $K_i$, and $f(A)_p = f_p(A_p)$, where $f_p$ is the induced map $A_p \rightarrow B_p$. We have directly that $$\{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}_p \subseteq \{b \in B_p: b \otimes 1 = 1 \otimes b \in B_p \otimes_{A_p} B_p\}$$where we use the canonical identification $(B \otimes_A B)_p \simeq B_p \otimes_{A_p} B_p$. By our earlier analysis (of $A$ just a field) then the result follows, as all of the RHS is $f_p(A_p)$.
• Thanks a lot. So the only important thing is that $A$ is a product of fields. For the more general question -- for what maps $f:A \to B$ does $f(A) = \{b \in B: b \otimes 1 = 1 \otimes b \in B \otimes_A B\}$ -- your last paragraph makes it clear that we can work locally. Do you have any idea when this happens more generally (or good examples for when it doesn't happen), other than $A$ a field? Dec 26 '12 at 20:23